Two bulbs are connected in parallel in the circuit, the total resistance of the two lamps is 8 ohm, the resistance of the bulb L1 is 40 ohm, and the total current of the trunk circuit is 0.75 a 2) Current through bulb L1 3) current through bulb L2 4) resistance of bulb L2

Two bulbs are connected in parallel in the circuit, the total resistance of the two lamps is 8 ohm, the resistance of the bulb L1 is 40 ohm, and the total current of the trunk circuit is 0.75 a 2) Current through bulb L1 3) current through bulb L2 4) resistance of bulb L2

1/R=1/R1+1/R2
1/8=1/40+1/R2
The solution is: R2 = 10 Ω
1) Power supply voltage: u = IR = 0.75 * 8 = 6V
2）I1=(1/5)*0.75A=0.15A
3）I2=(4/5)*0.75A=0.6A
4) R2 = 10 Ω

In the circuit shown in the figure, the range of ammeter is 0 ~ 0.6A, and that of voltmeter is 0 ~ 3V. The sliding rheostat is marked with "50 Ω 1A". When the switch S is closed and the slide of sliding rheostat slides to a certain position, the indication of ammeter is 0.3A, and that of voltmeter is 3V; when the slide of sliding rheostat slides to another position, the indication of ammeter is 0.5A, and that of voltmeter is 3V The indication is 1V. Calculate: (1) the resistance value and power supply voltage of resistor R1; (2) to ensure the safety of the circuit, adjust the range of sliding rheostat R2

(1) Switch S is closed, when the current number I1 = 0.3A, voltage number U1 = 3V, power supply voltage U = i1r1 + U1 = 0.3A × R1 + 3V ① When I2 = 0.5A, U2 = 1V, u = i2r1 + U2 = 0.5A × R1 + 1V ② The solution of the two equations is u = 6V, R1 = 10 Ω